1. The problem

A causal LTI system is described by the difference equation $\displaystyle y[n] - 5y[n-1] + 6y[n-2] = 2x[n-1]$.

Determine the

(a)homogeneous($\displaystyle x[n]=0$ for all $\displaystyle n$),

(b)impulse,

(c)step

responseof the system.

2. Relevant equations and definitions

2.1Causalare systems for which the output $\displaystyle y[n_0]$ depends only on the input samples $\displaystyle x[n]$, for $\displaystyle n \leq n_0$

2.2For the(a)part of the problem:

$\displaystyle \sum_{k=0}^{N} a_k y_h[n-k] = 0 \text{ ................... (2.96)}$

Equation $\displaystyle (2.96)$ is called thehomogeneous difference equationand $\displaystyle y_h[n]$ the homogeneous solution. The sequence $\displaystyle y_h[n]$ is in fact a member of a family of solutions of the form

$\displaystyle y_h[n] = \sum_{m=1}^{N} A_m z_m^{n} \text{ ................... (2.97)}$

where the coefficients $\displaystyle A_m$ can be chosen to satisfy a set of auxiliary conditions on $\displaystyle y[n]$.

3. The attempt at a solution

First of all, there is no additional information to be gained from knowing that the LTI system is also "causal", right? That is, being causal ($\displaystyle y[n_0]$ only depends on $\displaystyle x[n]$, $\displaystyle n \leq n_0$) is already in the given equation, correct?

(a)homogeneous response, y_h[n]

The official solution is given in the form of Eq. $\displaystyle (2.97)$: $\displaystyle y_h[n] = A_1(2)^n + A_2(3)^n$

I have no idea how one should arrive at this solution. Neither writing it out for particular $\displaystyle n$'s (and hoping to glimpse a pattern), nor writing the system in the general form of $\displaystyle \sum_{k=0}^N a_k y[n-k] = \sum_{m=0}^M b_m x[n-m]$ and then finding the relevant coefficients (N=2: $\displaystyle a_0=1, a_1=-5, a_2=6$; M=1: $\displaystyle b_0=0, b_1=2$), gives way toward the solution.

(b)impulse response, h[n]

I do get the official solution ($\displaystyle h[n] = 2(3^n - 2^n)u[n]$), but I'm exploiting the Z-transform, which is introduced only after this chapter ... so there must be other way without Z-transform:

$\displaystyle H(z) = \frac{Y(z)}{X(z)} = \frac{2z^{-1}}{1 - 5z^{-1} + 6z^{-2}} = 2\left(\frac{1}{1-3z^{-1}} - \frac{1}{1-2z^{-1}}\right) \Rightarrow h[n] = 2(3^n - 2^n)u[n]$

(c)step response, s[n]

As in the(a)part, I'm completely stuck here. The solution should be $\displaystyle s[n] = \left( -8(2)^{n-1} + 9(3)^{n-1} + 1 \right) u[n]$, which smells of Z-transform, but I'm also interested if/how could one do without it?

All I can think of for(c)is using the result from(b)(the impulse response), but I get stuck:

$\displaystyle s[n] = \sum_{k=-\infty}^{\infty} x[n-k] h[k]= \sum_{k=-\infty}^{\infty} u[n-k] 2(3^k - 2^k)u[k] = 2 \left( \sum_{k=-\infty}^{\infty} 3^k u[k] u[n-k] - \sum_{k=-\infty}^{\infty} 2^k u[k] u[n-k] \right) = \text{?} $